If the `k^(th)` term is the middle term in `(x^2 - (1)/(2x))^20` find
`T_k` and `T_(k+3)`

Answer

To solve the problem, we need to find the \( k^{th} \) term and the \( (k+3)^{th} \) term in the expansion of \( (x^2 - \frac{1}{2x})^{20} \), given that the \( k^{th} \) term is the middle term.

### Step-by-step Solution:

1. **Identify the Middle Term**:
The given expression is \( (x^2 - \frac{1}{2x})^{20} \). Since the exponent \( n = 20 \) is even, the middle term can be found using the formula:
\[ \text{Middle term} = T_{\left(\frac{n}{2} + 1\right)} = T_{11} \]

2. **Find \( T_{11} \)**:
The general term in a binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = x^2 \), \( b = -\frac{1}{2x} \), and \( n = 20 \). For \( T_{11} \), we have \( r = 10 \):
\[ T_{11} = \binom{20}{10} (x^2)^{20-10} \left(-\frac{1}{2x}\right)^{10} \]
Simplifying this:
\[ T_{11} = \binom{20}{10} (x^2)^{10} \left(-\frac{1}{2}\right)^{10} (x^{-10}) \]
\[ = \binom{20}{10} x^{20} \cdot \frac{(-1)^{10}}{2^{10}} \cdot x^{-10} \]
\[ = \binom{20}{10} \cdot \frac{1}{2^{10}} \cdot x^{10} \]

3. **Calculate \( T_{11} \)**:
\[ T_{11} = \binom{20}{10} \cdot \frac{1}{1024} \cdot x^{10} \]
The value of \( \binom{20}{10} \) is 184756, so:
\[ T_{11} = \frac{184756}{1024} x^{10} \]

4. **Find \( k \)**:
Since \( T_k \) is the middle term, we have \( k = 11 \).

5. **Find \( T_{k+3} = T_{14} \)**:
Now we need to find \( T_{14} \) (which is \( T_{k+3} \)):
\[ T_{14} = \binom{20}{13} (x^2)^{20-13} \left(-\frac{1}{2x}\right)^{13} \]
Here, \( r = 13 \):
\[ T_{14} = \binom{20}{13} (x^2)^{7} \left(-\frac{1}{2}\right)^{13} (x^{-13}) \]
\[ = \binom{20}{13} x^{14} \cdot \frac{(-1)^{13}}{2^{13}} \cdot x^{-13} \]
\[ = \binom{20}{13} \cdot \frac{-1}{8192} \cdot x^{1} \]
Since \( \binom{20}{13} = \binom{20}{7} = 77520 \):
\[ T_{14} = \frac{-77520}{8192} x \]

### Final Answers:
- \( T_k = T_{11} = \frac{184756}{1024} x^{10} \)
- \( T_{k+3} = T_{14} = \frac{-77520}{8192} x \)